Question 268: Magnetism and Matter

Question

An electron with kinetic energy $$5 \mathrm{~eV}$$ enters a region of uniform magnetic field of 3 $$\mu \mathrm{T}$$ perpendicular to its direction. An electric field $$\mathrm{E}$$ is applied perpendicular to the direction of velocity and magnetic field. The value of E, so that electron moves along the same path, is __________ $$\mathrm{NC}^{-1}$$.
(Given, mass of electron $$=9 	imes 10^{-31} \mathrm{~kg}$$, electric charge $$=1.6 	imes 10^{-19} \mathrm{C}$$)

Accepted Answer

To solve this problem, we first need to understand that we want the electron to move along the same path in the presence of both electric and magnetic fields. This implies that the forces due to the electric field and magnetic field must balance each other.
The force on an electron due to the electric field is given by:
$$F_E = eE$$
where $$e$$ is the charge of the electron and $$E$$ is the electric field.
The force on an electron due to the magnetic field (Lorentz force) is given by:
$$F_B = evB$$
where $$v$$ is the velocity of the electron and $$B$$ is the magnetic field.
For the electron to move in a straight path, the forces due to the electric field and magnetic field must be equal in magnitude:
$$eE = evB$$
From this, we can solve for the electric field $$E$$:
$$E = vB$$
Next, we need to find the velocity $$v$$ of the electron. The kinetic energy (KE) of the electron is related to its velocity by the equation:
$$KE = \frac{1}{2} mv^2$$
Given the kinetic energy (KE) is $$5 \, 	ext{eV}$$, we first convert this energy into joules since the given constants are in SI units:
$$5 \, 	ext{eV} = 5 	imes 1.6 	imes 10^{-19} \, 	ext{J} = 8 	imes 10^{-19} \, 	ext{J}$$
Now, solving for $$v$$:
$$8 	imes 10^{-19} = \frac{1}{2} 	imes 9 	imes 10^{-31} 	imes v^2$$
Rearrange to solve for $$v^2$$:
$$v^2 = \frac{2 	imes 8 	imes 10^{-19}}{9 	imes 10^{-31}}$$
$$v^2 = \frac{16 	imes 10^{-19}}{9 	imes 10^{-31}}$$
$$v^2 = \frac{16}{9} 	imes 10^{12}$$
$$v = \sqrt{\frac{16}{9} 	imes 10^{12}}$$
$$v = \frac{4}{3} 	imes 10^6 \, 	ext{m/s}$$
Now we can find the electric field $$E$$. Using the value of the magnetic field $$B$$ given as $$3 \, \mu 	ext{T} = 3 	imes 10^{-6} \, 	ext{T}$$:
$$E = vB = \left(\frac{4}{3} 	imes 10^6 \, 	ext{m/s}ight) 	imes \left(3 	imes 10^{-6} \, 	ext{T}ight)$$
$$E = \frac{4}{3} 	imes 3 	imes 10^0 \, 	ext{N/C}$$
$$E = 4 \, 	ext{N/C}$$
Therefore, the value of the electric field $$E$$ required for the electron to move along the same path is:
$$\boxed{4 \, 	ext{N/C}}$$

Magnetism and Matter

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